Find the value of $k$ if $\det(A^{–1}) = (\det A)^k$ where $A$ is a $3 \times 3$ matrix Now I'm getting two answers to this question i.e. $k = -1$ and $k = 1$
Reason i'm getting $k = 1$ :
$$A(A^{-1}) = I$$
Taking determinant on both sides we get
$$\det[A(A^{-1})] = \det(I)$$
$$\det(A)(\det(A^{-1})) = 1$$
Dividing both sides with $\det(A)$ we get
$$\det(A^{-1}) = \det(A)^{-1}$$
Hence, $k = -1$
Reason I'm getting $k = 1$ :
$$A^{-1} = \frac1{\det(A)}Adj(A)$$
Applying determinant on both sides we get
$$\det(A^{-1}) = \det{[\frac1{\det(A)}]Adj(A)}$$
$$\det(A^{-1}) = \frac1{\det(A)}\det[Adj(A)]$$
because $\det[Adj(A)] = \det(A)^{n-1}$
i.e. $\det[Adj(A)] = \det(A)^{3-1}$
Or, $$\det[Adj(A)] = \det(A)^2$$
Thus, $$\det(A^{-1}) = \frac1{\det(A)}[\det(A)^2]$$
Or, $$\det(A^{-1}) = \det(A)^1$$
Hence $k = 1$
Now my Question is which one is the right answer 
And why am I getting two answers, have i neglected a condition or what
Any kind of help will be much appreciated.
P.S. $k = -1$ is the answer given in all books I've come across.
 A: First the mistake:
$$\det(A^{-1}) = \frac{1}{\det(A)^\color{red}{3}}{\det[Adj(A)]}$$
because $\det[Adj(A)] = \det(A)^{n-1}$
i.e. $$\det[Adj(A)] = \det(A)^{3-1}$$
Or, $$\det[Adj(A)] = \det(A)^2$$
Thus, $$\det(A^{-1}) = \frac1{\det(A)^3}[\det(A)^2]=\frac{1}{\det(A)}$$
which is a known result.
Next, this is how I would approach the problem:
$$\det(A^{-1})= \det(A)^k$$
Hence we have $$1=\det(A)^{k+1}$$
If we want the equality to hold as long as $\det(A) \neq 0$, we require $k+1=0$.
Notice that I did not use the propery of the size of $A$.
A: Well. Observe that if $A$ is $3\times 3$ matrix then
\begin{align}
\det(c A)= c^3 \det(A). 
\end{align}
Example:  Consider
\begin{align}
\det
\begin{pmatrix}
c & 0 & 0\\
0 & c & 0\\
0 & 0 & c
\end{pmatrix}
= c\det
\begin{pmatrix}
1 & 0 & 0\\
0 & c & 0\\
0 & 0 & c
\end{pmatrix}
= c^2\det
\begin{pmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & c
\end{pmatrix}
= c^3\det
\begin{pmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{pmatrix}
\end{align}
A: Suppose that $A$ has the property $det(A)=1$, then $det(A^{-1})=1$, hence
$det(A^{-1})= ( \det(A))^k$  for all(!) $k \in \mathbb Z$.
